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Is it okay to apply divergence test on a series $\sum a_n$ and show that this series diverges by showing that $\lim \limits_{n \to \infty } |a_n| = \infty$?

If I have alternating sequence of $(-1^n) n^2$, can I say absolute value of this, $n^2$ goes to infinity therefore its series diverges by Divergence Test?

The definition of "$n$-th term Divergence Test" for the series $\sum a_n$ only mentions that I must show the divergence of the sequence $(a_n)$, not the divergence of the sequence $(|a_n|)$. I guess a better way of putting this question is to ask the following. Does the divergence of absolute value of the terms of a sequence imply the divergence of the original sequence?

Thanks in advance!

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    $\begingroup$ What do you mean by "divergence test"? There are many test for divergence of a series. Also recall that if $\sum a_n $ converges then $\lim a_n =0$. In other words, if the terms $a_n$ do not tend to $0$ then $\sum a_n$ diverges (note that the converse is not true). So stating that $\lim (-1)^n n^2 \neq 0$ suffices to conclude that the series diverges. $\endgroup$
    – Reveillark
    Commented Mar 28, 2015 at 3:37
  • $\begingroup$ I see that makes sense, thanks! That last conditional statement you made is what I meant by the divergence test. $\endgroup$
    – StatsNoob
    Commented Mar 28, 2015 at 3:51
  • $\begingroup$ Note that the converse of your 'question', however, is true: if $\sum a_n$ diverges, then $\sum |a_n|$ diverges. $\endgroup$
    – Leponzo
    Commented Mar 16, 2016 at 23:23
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    $\begingroup$ On a related note, if we consider the contrapositive of the theorem "$\sum \limits_n a_n ~ \text{converges} \implies \lim a_n=0$" we get "$\lim a_n \neq 0 \implies \sum \limits_n a_n ~ \text{ diverges}$". The question immediately can be asked, does this argument work for $\lim |a_n| $. Specifically is it true that $\lim |a_n| \neq 0 \implies \sum \limits_n a_n ~ \text{ diverges}$ ? $\endgroup$
    – john
    Commented Apr 21, 2022 at 19:51
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    $\begingroup$ @Reveillark does $ \lim |a_n| \neq 0 \implies \lim a_n \neq 0 $, as $n \to \infty $? I think that was the original question of the OP, but I am curious how we can prove it. Once we prove this, we get two implication chains: $ \lim |a_n| \neq 0 \implies \lim a_n \neq 0 \implies \sum \limits_n a_n ~ \text{ diverges} \implies \sum \limits_n |a_n| ~ \text{diverges} $ , and , $\sum \limits_n |a_n| ~ \text{converges} \implies \sum \limits_n a_n ~ \text{converges} \implies \lim a_n = 0$. $\endgroup$
    – john
    Commented Apr 21, 2022 at 20:11

1 Answer 1

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The series $\sum_{n \geq 1} \frac{(-1)^n}{n^2}$ converges because $(1/n^2)_{n \geq 1}$ is a sequence of positive terms that decreases to zero, c.f alternating series test.

If I understood right, the answer to your question is no: there exists series $\sum a_n$ which converge, but $\sum |a_n|$ diverges, for example the harmonic series $\sum (-1)^n/n$. These ones are called conditionally convergent. In fact, given such a series, you can reorder its terms so that the reordered series converge to whatever you want.

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    $\begingroup$ On a related note, if we consider the contrapositive of the theorem "$\sum \limits_n a_n ~ \text{converges} \implies \lim a_n=0$" we get "$\lim a_n \neq 0 \implies \sum \limits_n a_n ~ \text{ diverges}$". The question immediately can be asked, does this argument work for $\lim |a_n| $. Specifically is it true that $\lim |a_n| \neq 0 \implies \sum \limits_n a_n ~ \text{ diverges}$ ? $\endgroup$
    – john
    Commented Apr 21, 2022 at 19:46
  • $\begingroup$ Can we say that $\lim|{a_n}| \neq 0$ implies $\lim{a_n} \neq 0$? Cause if $a_n$ is negative and it's not approaching 0 then it's only -1 times the term. $\endgroup$
    – Nabs
    Commented Aug 13, 2023 at 0:09

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